Question

Hi ♡

Question :
A three digit number is equal to 17 times the sum of the digits. If the digits are reversed the new number is 198 more than the original number. The sum of the extreme digits is 1 less than the middle digit. Find the original number.

Please solve using just two variables.

Thank you! ♥​

Answer 1

HeLLo❤!!Manu Here xD!

Solution:Let the hundreds place digit to be X,tens digit be Y and unit digit to be Z.

* So, 100x+10y+Z

So, 100x+10y+ZIn Question,we are provided that A three digit number is equal to 17 times the sum of its digit.

100x+10y+Z=17(x+y+z)

100x+10y+z=17x+17y+17z

y

100x-17x+10y-17y+z-17z=17x-83x-7y-16z=0______(1)

____________________

* If the digits are reversed the new number is 198 more than the original number.

100x+10y+z+198=100z+10y+x

100x-x+10y-10y+z-100z+198=99x-99z+198=0

z-x=2

z=x+2__________(2)

*The sum of the extreme digits is 1 less than the middle digit.

And,

z+x=y-1

By Putting (2)

z+z-2=y-1

Y=2z-1

__________(3)

BY PUTTING (3),(2)AND (1),WE GET:

83( z- 2 ) -7( 2z -1 )- 16z =0

z=3

Put,z=3 in ( 3 )

Y=2 ( 3 ) - 1

Y =5

Now,Put z = 3 in ( 2 )

x=3-2 = 1

So,Number =100 ( 1 ) + 10 ( 5 ) + 3= 153

___________________________________

Thanks❤!

Answer 2

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

A three digit number is equal to 17 times the sum of the digits. If the digits are reversed the new number is 198 more than the original number. The sum of the extreme digits is 1 less than the middle digit. Find the original number?

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• A three digit number is equal to 17 times the sum of the digits.
• The sum of the extreme digits is 1 less than the middle digit.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

Considering the Third statement,

The sum of the extreme digits is 1 less than the middle digit.

Let the Extreme digits be x and y.

From the statement,

⇒ x + y = c - 1

⇒ c = x + y + 1.

Therefore the Middle digit is x + y + 1.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, Let's assume the number be,

N = 100x + 10(x + y + 1) + y

N = 100x + 10x + 10y + 10 + y

N = 110x + 11y + 10

where,

• x be the digit in Hundreds place.
• (x + y + 1) be the digit in tens place.
• y be the digit in Units place.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Considering Statement- 1

A three digit number is equal to 17 times the sum of the digits.

⇒ N = 17 × (Sum of the all digits)

⇒110x + 11y + 10= 17 × (x + [x + y + 1]+ y)

⇒110x + 11y + 10 = 17 × (x + x + y + 1 + y)

⇒110x + 11y + 10 = 17 × (2x + 2y + 1)

⇒110x + 11y + 10 = 34x + 34y + 17

⇒110x + 11y + 10 = 34x + 34y + 17

⇒ (110x - 34x) + (11y - 34y) + 10 - 17 = 0

⇒ 76x - 23y - 7 = 0 ---------(1)

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Considering Statement- 2

If the digits are reversed the new number is 198 more than the original number.

When we Reverse the Number we get,

N' = 100y + 10(x + y + 1) + x

N' = 100y + 10x + 10y + 10 + x

N' = 110y + 11x + 10

Formulating the Equation,

⇒ N +198 = N'

⇒110x + 11y + 10 + 198 = 110y + 11x +10

⇒(110x - 11x)+(11y - 110y)+10-10+198= 0

⇒ 99x - 99y + 198 = 0

⇒ 99 (x - y + 2) = 0

⇒ x - y + 2 = 0

⇒ x = y - 2 --------(2)

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Substituting Equation (2) in Equation (1).

76x - 23y - 7 = 0 ---------(1)

Substituting,

⇒ 76(y - 2) - 23y - 7 = 0

⇒ 76y - 152 - 23y - 7 = 0

⇒ (76y - 23y) - 152 - 7 = 0

⇒ 53y - 159 = 0

⇒ 53y = 159

⇒ y = 159/53

⇒ y = 3.

Substituting y = 3 in equation (2).

x = y - 2 --------(2)

⇒ x = 3 - 2

⇒ x = 1.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, Calculating the Number,

N = 100x + 10(x + y + 1) + y

Substituting the values,

⇒ N = 100 × 1 + 10(1 + 3 + 1) + 3

⇒ N =100 + 10 × 5 + 3

⇒ N = 100 + 50 + 3

⇒ N = 153.

So, the Number is 153.

$\rule{300}{1.5}$